Application of DerivativeHard
Question
The set of values of p for which the extremum of the function f(x) = x3 - 3 px2 + 3 (p2 - 1) x + 1 lie in the interval (- 2, 4), is:
Options
A.(- 3, 5)
B.(- 3, 3)
C.(- 1, 3)
D.(- 1, 4)
Solution
f′(x) = 3x2 - 3p2x + 3p2 - 3
= 3((x - p)2 - 1)
= 3(x - (p + 1)) (x - (p - 1))
⇒ p - 1 > - 2 and p + 1 < 4
⇒ p > - 1 and p < 3
⇒ - 1 < p < 3
= 3((x - p)2 - 1)
= 3(x - (p + 1)) (x - (p - 1))
⇒ p - 1 > - 2 and p + 1 < 4
⇒ p > - 1 and p < 3
⇒ - 1 < p < 3
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